Download Commutative Algebra Notes Introduction to Commutative Algebra

Commutative Algebra Notes
Introduction to Commutative Algebra
Atiyah & Macdonald
Adam Boocher
1
Rings and Ideals
1.1
Rings and Ring Homomorphisms
A commutative ring A with identity is a set with two binary operations (addition
and multiplication so that for all x, y, z ∈ A:
1. A is an abelian group with respect to addition (so it contains a zero element, 0, and every x ∈ A has an additive inverse −x.
2. Multiplication is associative ((xy)z = x(yz)) and distributive over addition
(x(y + z) = xy + xz)
3. Multiplication is commutative (xy = yx)
4. There is an identity element 1 with x1 = 1x = x
The identity element is unique.
Comment. If 1 = 0 then for any x ∈ A,
x = x1 = x0 = 0
and we call A the zero ring denoted by 0.
A ring homomorphism is a mapping f of a ring A into a ring B such that
for all x, y ∈ A, f (x + y) = f (x) + f (y), f (xy) = f (x)f (y) and f (1) = 1. The
usual properties of ring homomorphisms can be proven from these facts.
A subset S of A is a subring of A if S is closed under addition and multiplication and contains the identity element of A. The identity inclusion map
f : S → A is then a ring homomorphism.
The composition of two homomorphisms is a homomorphism.
1
1.2
Ideals, Quotient Rings
A subset a of A is an ideal of A if a is closed under addition and Aa = aA = a.
(Meaning that ra ∈ a for all r ∈ A, a ∈ a). The quotient group A/a is then a
ring by the obvious multiplication (a + a)(b + a) = (ab + a). We call this ring
the quotient ring A/a. The map π : A → A/a defined by π(x) = x + a is a
surjective ring homomorphism.
Proposition 1.1. There is a 1-1 correspondence between the ideals b containing
a and the ideals b of A/a.
Proof. We prove this with careful detail. Let b be an ideal containing a. Then
define φ(b) = {x + a : x ∈ b}, an ideal in A/a. Conversely let b be an ideal of
A/a. Then take ψ(b) = {x ∈ A : x + a ∈ b}. This is an ideal of A and contains
a. If we can check that φ ◦ ψ = id and ψ ◦ φ = id then we are done. This is easy,
though. For the first, Let b be an ideal of A/a. Then ψ(b) = {x ∈ A : x+a ∈ b}.
Then
φ(ψ(b)) = {x + a : x ∈ ψ(b)} = {x + a : x ∈ {x ∈ A : x + a ∈ b}}
which translates to saying that φ(ψ(b)) = {x + a : x + a ∈ b}. So that φ(ψ(b)) =
b. The other composition is similar.
If f : A → B is any ring homomorphism, the kernel of f , that is is the set
of all a ∈ A so that f (a) = 0 is an ideal of A and the image of f is a subring of
B, and f induces a ring isomorphism
A/kerf ∼
= Imf.
Which you might like to call the first isomorphism theorem for rings. The
notation x ≡ y (mod a) means that x − y ∈ a.
1.3
Zero-Divisors, Nilpotent Elements, Units
A zero-divisor in a ring A is an element x so that there exists a nonzero element
y ∈ A with xy = 0. A ring with no nonzero zero-divisors is called an integral
domain.
An element x ∈ A is nilpotent if there exists and n > 0 so that xn = 0. A
nilpotent element is a zero-divisor (unless A = 0) since xn−1 x = 0.
A unit in A is an element x so that there exists an element y ∈ A with
xy = 1. This y is determined uniquely and is denoted x−1 . The units of A form
a multiplicative group.
The multiples ax of an element x ∈ A is called the principal ideal generated
by x and is denoted (x). Note that x is a unit iff (x) = A = (1).
A field is a ring A in which 1 6= 0 and every non-zero element is a unit.
Every field is an integral domain. (If xy = 0, then y = x−1 xy = 0).
Proposition 1.2. Let A be a nonzero ring. Then the following are equivalent:
2
1. A is a field
2. the only ideals in A are 0 and (1);
3. every homomorphism of A into a non-zero ring B is injective.
Proof. 1 =⇒ 2: Let a be a nonzero ideal with x ∈ a nonzero. Then x is a unit
and thus a ⊇ (x) = (1).
2 =⇒ 3. Let f : A → B be a homomorphism and B nonzero. Then kerf is
an ideal of A and is either (1) or 0. If it is (1) then f = 0 which is impossible
since f (1) = 1. Thus kerf = 0 and f is injective.
3 =⇒ 2. Let x be a nonzero element of A. Suppose that x is not a unit so
that (x) is not equal to (1). Then A/(x) is a nonzero ideal and hence the natural
homomorphism π : A → A/(x) is injective. But this means that (x) =ker(f ) = 0
which is a contradiction.
1.4
Prime Ideals and Maximal Ideals
An ideal p in A is prime if p 6= (1) and if xy ∈ p then either x ∈ p or y ∈ p. An
ideal m in A is maximal if m 6= (1) and if there is no ideal a such that
m ⊂ a ⊂ (1)
(strict inclusions).
Equivalently we have the following mini-proposition/definition:
p is prime ⇐⇒ A/p is an integral domain;
m is maximal ⇐⇒ A/m is a field
Should we prove this? This first is quite easy, so we’ll work out the second. Let
m be maximal. Then since there is a correspondence between ideals of A/m
and ideals containing m, the maximality of m says that A/m has no non-trivial
ideals and Proposition 1.2 thus guarantees that A/m is a field. Conversely, if
A/m is a field, then by the correspondence again, there are no ideals between
m and A.
Comment. Note that this implies that maximal ideals are prime, but not necessarily vice versa. Also note that the zero ideal is prime ⇐⇒ A is an integral
domain.
Proposition 1.3. If f : A → B is a ring homomorphism (henceforth abbrev.
HM) and b is a prime ideal in B then f −1 (b) is a prime ideal in A.
Proof. This is very direct. The fact that f −1 (b) is an ideal is immediate. Let
xy ∈ f −1 (b). Then f (xy) ∈ b and therefore f (x)f (y) ∈ b it follows since b is
prime that either f (x) or f (y) ∈ b and thus either x or y ∈ f −1 (b).
Comment. The corresponding statement about maximal ideals is not true in
general, since if A is any ring that is not a field and F is any field, then 0 is
maximal in F but its inverse image in A may not be maximal at all. (Just let
f be injective!)
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Theorem 1.1. Every ring A 6= 0 has at least one maximal ideal.
The proof will use Zorn’s Lemma so remind you first what it says. Let S be
a partially ordered set (sometimes called a poset), that is, one with a relation
≤ that is reflexive, transitive and anti-symmetric. A subset T of S is called a
chain if any two elements of T are comparable. That is to say that if x, y ∈ T
then either x ≤ y or y ≤ x. An upper bound for a T in S is an element x ∈ S
such that t ≤ x for every t ∈ T . Finally, a maximal element in S is an element
x ∈ S so that for all y such that x ≤ y, we have x = y.
Theorem 1.2 (Zorn’s Lemma). If every chain T of S has an upper bound in
S then S has at least one maximal element.
Zorn’s Lemma is equivalent to the axiom of choice.
Proof. (of Theorem) Let Σ be the set of all ideals not equal to (1) in A. Order
Σ by inclusion. Σ is not empty, since 0 ∈ Σ. We must show that every chain
in Σ has an upper bound in Σ. Thus we are inspired to let (aα ) be a chain of
ideals in Σ, so that for each pair of indices α, β we have either
aα ⊆ aβ
or aβ ⊆ aα .
S
Let a = α aα . We claim that a is an ideal. Indeed, a is clearly closed under
multiplication by A, so we show closure under addition. Let x, y ∈ a. Then
x ∈ aα , y ∈ aβ for some α, β. Then one of these ideals contains the other, since
they are elements of a chain and we therefore have x, y contained in the same
ideal and thus x + y ∈ a. Note that 1 ∈
/ a since 1 ∈
/ aα for all α. Hence a ∈ Σ
and a is an upper bound of the chain. Thus by Zorn’s Lemma, Σ contains a
maximal element. What is a maximal element in Σ? It is an ideal that does
not contain 1 so that it there is larger ideal in Σ containing it; a maximal ideal
in A.
Corollary 1.1. If a 6= (1) is an ideal of A, there exists a maximal ideal of A
containing a.
Proof. Note that A/a has a maximal ideal m by the above theorem. Denote
by n the corresponding ideal of A containing a. We claim that n is maximal
in A. The claim is justified as follows: suppose that there is an ideal p strictly
between (1) and n. Then the ideal p = {x + a : x ∈ p} =
6 (1) is an ideal of A/a
that strictly contains m which is a contradiction.
Corollary 1.2. Every non-unit of A is contained in a maximal ideal. (Just let
a = (x)).
Comment. There exist rings with exactly one maximal ideal, for example fields.
A ring A with exactly one maximal ideal m is called a local ring and the field
k = A/m is called the residue field of A.
Proposition 1.4. Let A be a ring and m 6= (1) and ideal of A such that every
x ∈ A − m is a unit in A. Then A is a local ring and m its maximal ideal.
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Proof. Note that m is clearly maximal in A since if an ideal contained m and
any other element, it would contain a unit. Also, all ideals 6= (1) consist of nonunits. Note that m contains all non-units so that m contains all ideals 6= (1),
and is therefore the only maximal ideal.
Proposition 1.5. Let A be a ring with m a maximal ideal of A, such that every
element of 1 + m is a unit in A. Then A is a local ring.
Proof. Let x ∈ A − m. Then the ideal generated by m and x is (1) since m is
maximal. It follows then that there exist m ∈ m and t ∈ A so that m + xt = 1.
This implies that
xt = 1 − m ∈ 1 + m
and thus xt is a unit. This also implies that x is a unit. (Do you see why?) Thus
we can apply the previous proposition and conclude that A is a local ring.
A ring with only finitely many maximal ideals is called semi-local.
Example 1.1. See Atiyah & MacDonald for examples.
Proposition 1.6. A principal ideal domain (PID) is an integral domain A
where every ideal is principle. In such a domain, every nonzero prime ideal is
maximal.
Proof. Suppose that (x) is a prime ideal, but not maximal. Let (y) ⊃ (x). Then
since (x) sits inside of (y), we see that x = yt for some t ∈ A. Thus yt ∈ (x)
and y ∈
/ (x). Thus it follows that t ∈ (x) since (x) is prime and hence t = xw
for some w ∈ A. Substituting this, we see that
x = yt = yxw =⇒ yw = 1
and y is a unit so that (y) = (1) and (x) is maximal.
1.5
Nilradical and Jacobson Radical
Proposition 1.7. The set R of all nilpotent elements in a ring A is an ideal
and A/R has no nilpotent elements 6= 0.
Proof. If x is nilpotent then so is ax for all a ∈ A. Now let x, y be nilpotent
elements, say xn , y m = 0. Then
(x + y)n+m−1 = 0
since each term of the expansion must contain either a power of x greater than
n − 1 or a power of y greater than m − 1. To see that A/R has no nilpotent
elements, note that x + R ∈ A/R is nilpotent if and only if xn + R = 0 in A/R
which is equivalent to saying that xn ∈ R which would imply that x ∈ R.
The ideal R defined above is called the nilradical of A. The following proposition gives another definition of R.
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Proposition 1.8. The nilradical of A is the intersection of all the prime ideals.
Proof. Let R0 denote the intersection all prime ideals of A. Then if f is nilpotent, then f n = 0 for some n > 0. Since 0 ∈ p for all ideals and p is prime, we
have that f ∈ p for all prime ideals p and hence f ∈ R0 .
Conversely, we will show that if f is not nilpotent, then it is not in the
intersection of all prime ideals. Suppose that f is not nilpotent. Then let Σ be
the set of all ideals a such that no power of f is in a. Ordering Σ by inclusion
we can apply Zorn’s Lemma to conclude that it has a maximal element, p. We
shall show that p is prime by showing that x, y ∈
/ p implies xy ∈
/ p. Indeed, if
x, y ∈
/ p then p + (x) and p + (y) properly contain p and thus are not elements
of Σ by the maximality of Σ. Thus it follows that there exist some n, m so that
f n ∈ p + (x)
f m ∈ p + (y)
which clearly imply that f n+m ∈ p + (xy) which implies that xy ∈
/ p. Thus p is
prime and does not contain f as required.
The Jacobson radical R of A is the intersection of all the maximal ideals of
A. It can be characterized as follows:
Proposition 1.9. x ∈ R ⇐⇒ 1 − xy is a unit in A for all y ∈ A.
Proof. ⇒: Suppose 1 − xy is not a unit for some y ∈ A. Then by Corollary 1.2
1 − xy is contained in some maximal ideal m of A. But since x ∈ R ⊂ m we
have 1 − xy ∈ m =⇒ 1 ∈ m which is absurd.
⇐: Suppose x ∈
/ m for some maximal ideal m. Since x and m generate A
we have m + xy = 1 for some elements m ∈ m and y ∈ A. Thus 1 − xy ∈ m
contradicting the fact that 1 − xy is a unit.
2
Operations of Ideals
We define the sum of two ideals a and b to be the the set of all x + y where x ∈ a
and y ∈ b. We denote the sum by aP
+ b. It is the smallest ideal containing a
and b. In general we define the sum i∈I
P ai of any family (possibly infinite) of
ideals ai of A; its elements are all sums
xi , where xi ∈ ai for all i ∈ I and all
but finitely many of the xi are zero. It is the smallest ideal of A which contains
all of the ideals ai .
The intersection of any family of ideals is an ideal. (Proof is easy)
The product of two ideals a, b in A is the ideal generated
by all products xy
P
where x ∈ a and y ∈ b. It is the set of all finite sums
xi yi where xi ∈ a and
yi ∈ b. We can similarly define the product of any finite family of ideals. In
particular, the powers an of an ideal are defined. We also make the convention
that a0 = (1).
Example 2.1.
6
1. If A = Z, a = (m), b = (n) then a + b is the ideal generated by gcd(m, n).
This follows since the gcd of any two numbers can always be represented
as an integer combination of the two numbers; a ∩ b is the ideal generated
by their lcm (proof easy); and ab = (mn) (proof easy) Thus it follows that
ab = a ∩ b ⇐⇒ m, n are coprime.
2. A = k[x1 , . . . , xn ], a = (x1 , . . . , xn ) = ideal generated by (x1 , . . . , xn ).
Then am is the set of all polynomials with no terms of degree < m.
The above operations are all commutative and associative. We also have a
distributive law for products.
a(b + c) = ab + ac
In general, ∩ and + are not distributive over each other. The best we can do is
the modular law
a ∩ (b + c) = a ∩ b + a ∩ c if a ⊇ b or a ⊇ c.
Indeed, if a ⊇ b and x ∈ a ∩ (b + c) then x = b + c for some b ∈ b ⊆ a and c ∈ c.
Since x ∈ a we have b + c ∈ a and thus that c ∈ a. The rest is easy to do.
Finally, (a + b)(a ∩ b) ⊆ ab. (since (a + b)(a ∩ b) = (a(a ∩ b) + b(a ∩ b) ⊆ ab).
Also we clearly have that ab ⊆ a ∩ b, hence
a ∩ b = ab provided a + b = (1).
We say that two ideals a, b are coprime if a + b = (1). Thus for coprime ideals
we have a ∩ b = ab. Two ideals are coprime iff there exist x ∈ a and y ∈ b such
that x + y = 1.
Let A1 , . . . , An be rings. Their direct product
A=
n
Y
Ai
i=1
is the set of all sequences x = (x1 , . . . , xn ) with xi ∈ Ai with componentwise
addition and multiplication. A is a commutative ring with identity (1, . . . , 1).
We have projections pi : A → Ai defined by pi (x) = xi ; they are ring homomorphisms.
Let A be a ring and a1 , . . . , an ideals of A. Define a homomorphism
φ:A→
n
Y
(A/ai )
i=1
by the rule φ(x) = (x + a1 , . . . , x + an ).
Proposition 2.1.
i. If ai , aj are coprime whenever i 6= j, then Πai =
ii. φ is surjective ⇐⇒ ai , aj are coprime whenever i 6= j.
7
T
ai .
iii. φ is injective ⇐⇒
T
ai = (0).
Proof. i). We will induct on n, the number of ideals. When n = 2 the argument
is handled above. Suppose
we haveT
a1 , . . . , an and the result is true for any set
Qn−1
n−1
of n − 1 ideals. Let b = i=1 ai = i=1 ai (by assumption). Since ai + an = 1
for each i we know that we have equations
xi + yi = 1
Note that
n−1
Y
xi =
i=1
with xi ∈ ai , yi ∈ an
n−1
Y
(1 − yi ) ≡ 1(mod an ).
i=1
Hence an + b = (1) so they are coprime. Thus
n
Y
ai = ban = b ∩ an =
i=1
n
\
an .
i=1
ii). ⇒: We will show, for example, that a1 and a2 are coprime. We accomplish this by asserting noting that since φ is surjective, there is an x so
that
φ(x) = (1 + a1 , 0, . . . , 0).
This means that x ≡ 1 mod a1 and x ≡ 0 mod a2 . This means that
1 = (1 − x) + x ∈ a1 + a2
so that a1 and a2 are coprime.
⇐: It is enough to show that we can obtain (1 + a1 , 0, . . . , 0) for example.
Since a1 , ai are coprime for i > 1 we have a set of equations
ci + yi = 1
Let x =
Qn
i=2
where
ci ∈ a1 , yi ∈ ai .
yi . Then x is 0 mod ai when i > 2. On the other hand,
x=
n
Y
(1 − ci )
i=2
which is 1 mod a1 . TThus φ(x) = (1 + a1 , 0, . . . , 0).
iii). Note that ai = ker(φ) which obviously implies the result.
In general, the union of two ideals is not an ideal. The reason for this is that
the sum of two elements need not be contained in the ideal. For an example,
consider, Z with the union of the ideals (3), (5). This is the set of all integers
which are multiples of either 3 or 5. It is not closed under addition since 3+5 = 8
is not in the union. We can do better than that, however, with the following
“Prime Avoidance Lemma”.
8
Proposition 2.2. (Prime Avoidance Lemma)
Sn
Let p1 , . . . , pn be prime ideals and let a be an ideal contained in i=1 pi .
Then a ⊆ pi for some i.
Proof. This is a somewhat complicated argument so we first walk through the
steps. We would like to show that
a⊆
n
[
pi =⇒ a ⊆ pi for some i.
i=1
We will accomplish this by proving the contrapositive:
a * pi for all i =⇒ a *
n
[
pi .
i=1
We will do this by induction in fact! The statement is certainly true when n is
1. We now prove it for n > 0. Assume that the result is true for n − 1. Suppose
a * pi for all i = 1 . . . n. Then for each i the remaining n − 1 ideals satisfy the
induction hypothesis so we can say
a*
n
[
pj \pi .
j=1
This is the same as saying that there is an element xi ∈ a so that xi is not in
any of the pj when j 6= i. If for any i we have that xi ∈
/ pi then we will have
succeeded in showing the statement for n and are through. Thus suppose that
xi ∈ pi for all i. Consider the element
y=
n
X
x1 x2 · · · xi−1 xi+1 x1+2 · · · xn .
i=1
Then y ∈ a (clearly) and y ∈
/ pi (1 ≤ i ≤ n) (since y is the sum of n − 1 terms
in pi and one term not contained S
in pi . In fact we need the primeness of pi to
n
assert this latter part). Thus a * i=1 pi .
Proposition 2.3. Let a1 , . . . , an be ideals and let pTbe a prime ideal containing
their intersection. Then p ⊇ ai for some i. If p = ai , then p = ai for some i.
Proof. This proof is straightforward. Suppose the statement
isQfalse. T
Then for
Q
each i there exists and xi ∈ ai such that xi ∈
/ p. Then xi ∈ ai ⊆ ai ⊆ p.
But since p is prime, the product cannot beTan element of p. Thus we have a
contradiction and p ⊇ ai for some ai . If p = ai then p ⊆ ai for all i and hence
p = ai for some i.
If a, b are two ideals in a ring A, their ideal quotient is
(a : b) = {x ∈ A : xb ⊆ a}
9
This is an ideal: (This is clear) In particular, (0 : b) is called the annihilator of
b and is also denoted Ann(b): it is the set of all x ∈ A such that xb = 0. In this
notation, the set of all zero-divisors in A is
[
D=
Ann(x)
x6=0
where (x) is the principal ideal generated by x. In fact, if b = (x) is a principal
ideal, we will just use (a : x) instead of the fancier (a : (x)).
Example 2.2. Let A = Z, a = (m), b = (n), where
m = 2µ2 3µ3 5µ5 · · ·
,
n = 2ν2 3ν3 5ν5 · · ·
Then (a : b) is the set of all numbers which when multiplied with n give a
multiple of m. It is clear that (a : b) = (q) where
q = 2γ2 3γ3 5γ5 · · ·
and γp = max(µp − νp , 0). In other words, q = m/ gcd(m, n).
Exercise 1.12 i) a ⊆ (a : b). (Clear)
ii) (a : b)b ⊆ a. If x ∈ (a : b) then xb ⊆ a which is what this sentence is saying.
iii) ((a : b) : c) = (a : bc) = ((a : c) : b). This argument is done by noticing that
x ∈ ((a : b) : c) =⇒ xc ⊆ (a : b)
=⇒ xcb ⊆ a =⇒ x(bc) ⊆ a =⇒ x ∈ (a : bc)
All T
the other directions
are the same.
T
iv)( i aP
:
b)
=
(a
:
b).
These are both easyPto see.
i
i i
T
v)(a : i bi ) = i (a : bi ). Suppose
that x( i bi ) ⊆ a. Then
T
T in particular,
xbP
⊆
a
for
each
i.
Thus
x
∈
(a
:
b
).
Conversely,
if
x
∈
i
i
i
i (a : bi ), then
x( i bi ) ⊆ a as required.
If a is any ideal of A, the radical of a is
r(a) = {x ∈ A : xn ∈ a for some n > 0}
If φ : A → A/a is the standard homomorphism, then r(a) = φ−1 (RA/a ) (the
nilradical of the quotient). Thus since it is the preimage of an ideal, it is an
ideal.
Exercise 1.13 i) r(a) ⊇ a. (Clear)
ii) r(r(a)) = r(a).
Suppose x ∈ r(r(a)). Then for some n, xn ∈ r(a) and thus for some m,
n m
(x ) = xnm ∈ a. The other inclusion follows from i).
iii) r(ab) = r(a ∩ b) = r(a) ∩ r(b).
We prove the first inequality. Suppose that x ∈ r(ab). Then for some n,
xn ∈ ab ⊆ a ∩ b so x ∈ r(a ∩ b). Conversely, suppose x ∈ r(a ∩ b). Then for
some n, xn ∈ a, xn ∈ b which implies that x2n ∈ ab so x ∈ r(ab).
10
Now we prove the second equality. Let x ∈ r(a ∩ b). Then for some n,
xn ∈ a ∩ b ⊆ r(a) ∩ r(b) by i). Thus xn ∈ r(a) and xn ∈ r(b), giving us that in
fact x ∈ r(a) ∩ r(b). Conversely, if x ∈ r(a) ∩ r(b) then for some m, n, xm ∈ a,
xn ∈ b and thus xmn ∈ a ∩ b.
iv) r(a) = (1) ⇐⇒ a = (1).
Suppose that r(a) = (1). Then 1 ∈ r(a) so that 1 = 1n ∈ a, so a = (1). The
other direction is clear.
v)r(a + b) = r(r(a) + r(b)).
Let x ∈ r(a + b). Then for some n, xn ∈ a + b ⊆ r(a) + r(b) so that x ∈
r(r(a)+r(b)). Conversely, if x ∈ r(r(a)+r(b)) then for some n, xn ∈ r(a)+r(b).
This means xn = p + q where pm ∈ a and q k ∈ b for some m, k. Then for a
large enough power M , xM ∈ a + b and x ∈ r(a + b).
vi) if p is prime, r(pn ) = p for all n > 0.
First note that p is clearly contained in r(pn ). Now suppose x ∈ r(pn ). Then
for some m, xm ∈ pn ⊂ p. Thus since p is prime, x ∈ p.
Proposition 2.4. The radical of an ideal a is the intersection of the prime
ideals which contain a.
Proof. Proposition 1.8 applied to A/a tells us that the nilradical of A/a is the
intersection of all prime ideals of A/a which is in correspondence with the set
of all prime ideals containing a.
More generally, we may define the radical r(E) of any subset E of A in the
same way. It is not an ideal in general. We have
[
[
r( Eα ) =
r(Eα )
α
α
for any family of subsets Eα of A. (This is seemingly vacuous)
S
Proposition 2.5. D = set of zero-divisors of A = x6=0 r(Ann(x)).
S
S
Proof. D = r(D) = r( x6=0 Ann(x)) = x6=0 r(Ann(x)).
Example 2.3. If A = Z, a = (m),Tlet pi (1 ≤ i ≤ r) be the distinct prime divisors
r
of m. Then r(a) = (p1 · · · pr ) = i=1 (pi ).
Proposition 2.6. Let a, b be ideals in a ring A such that r(a), r(b) are coprime.
Then a, b are coprime.
Proof. Suppose that r(a) + r(b) = (1). Then by v) above,
r(a + b) = r(r(a) + r(b)) = r(1) = (1)
and by iv) we have that a + b = (1).
11
3
Extension and Contraction
Let f : A → B be a ring homomorphism. If a is an ideal in A, in general, f (a)
is not an ideal in B. (take for example, the imbedding of Z into Q and take a
to be any nonzero ideal in Z. We define the extension ae of a to be the ideal
Bf (a) generated by f (a) in B: explicitly, ae is the set of all sums
X
yi f (xi )
xi ∈ a ,
yi ∈ B.
If b is an ideal in B, then f −1 (b) is always an ideal of A called the contraction
bc of b. If b is prime, then bc is prime. (This is an easy proof). If a is prime in
A then ae is not necessarily prime in B. (Just inject Z into Q).
We can factorize this process as
p
j
A → f (A) → B
where p is surjective and j is an injection. For p the situation is very simple.
There is a bijective correspondence between ideals of f (A) and the ideals of
A which contain ker(f ). Let’s see why this is true. We know that f (A) ∼
=
A/ ker f (A). Now just apply (1.1). Prime ideals correspond to prime ideals.
For j, the situation is much more dicey. (Insert example from algebraic number
theory which is presented without proof in AM). Moving on...
Proposition 3.1. Let f : A → B be a ring homomorphism and a and b as
before. Then
i.) a ⊆ aec , b ⊇ bce ;
ii.) bc = bcec , ae = aece ;
iii.) If C is the set of contracted ideals in A and if E is the set of extended
ideals in B, then C = {a|aec = a}, E = {b|bce = b}, and a 7→ ae is a
bijective map of C onto E, whose inverse is b 7→ bc .
Proof. i.) Let x ∈ a. Then x ∈ f −1 (ae ) so x ∈ aec . Next let y ∈ bce . Then
y is a linear combination of terms of the form bi f (xi ) with bi ∈ B and
xi ∈ bc . This means that f (xi ) ∈ b and thus y ∈ b.
ii.) These both really do follow from i). Indeed we have bc ⊆ bcec by the first
part, and since the second part says b ⊇ bce we take the contraction of
both sides to conclude bc ⊇ bcec . The second equality is obtained in the
same manner.
iii.) Let a ∈ C. Then a is a contracted ideal so a = bc for some ideal b of B.
Then aec = bcec = bc = a. The fact that such a with this property are in
C follows since a = aec guarantees that a is a contraction. The proof for
E is nearly identical. To prove that the map a 7→ ae is bijective we note
that it is surjective since if ae ∈ E then ae = aece which is the extension
of the contraction of an ideal. To see that it is injective, suppose that
12
ce
cec
bce
= bcec
which
1 = b2 . Take the contraction of both sides to obtain b1
2
c
c
by ii). implies b1 = b2 as expected.
AM feels the need here to include a ridiculous number of properties of e and
c so we list them and maybe we’ll prove the hard ones sometime.
Proposition 3.2. If a1 , a2 are ideals of A and b1 , b2 are ideals of B then
(a1 + a2 )e = ae1 + ae2 ,
(a1 ∩ a2 )e ⊆ ae1 ∩ ae2 ,
(a1 a2 )e = ae1 ae2 ,
(a1 : a2 )e ⊆ (ae1 : ae2 ),
r(a)e ⊆ r(ae ),
(b1 + b2 )c ⊇ bc1 + bc2 ,
(b1 ∩ b2 )c = bc1 ∩ bc2 ,
(b1 b2 )c ⊇ bc1 bc2 ,
(b1 : b2 )c ⊆ (bc1 : bc2 ),
r(b)c = r(bc ).
The set of ideals E is closed under sum and product, and C is closed under the
other operations.
On second thought, I just worked all of these out and they are completely
trivial and are not difficult at all.
4
Exercises 1
1. Let x be a nilpotent element of a ring A. Show that 1 + x is a unit of A.
Deduce that the sum of a nilpotent element and a unit is a unit.
Suppose that xn = 0. Then wlog, let n be even (take n + 1 if you like) and
then (1 + x)(1 − x + x2 − · · · − xn−1 ) = 1 − xn = 1 so (1 + x) is a unit. Now
let c be a unit. Then (c + x) = c(1 + x/c) which is a product of two units since
x/c is nilpotent if x is.
2. Let A be a ring and let A[x] be the ring of polynomials in an indeterminate
x, with coefficients in A. Let f = a0 + a1 x + · · · + an xn ∈ A[x]. Prove that
i.) f is a unit in A[x] ⇐⇒ a0 is a unit in A and a1 , . . . , an are nilpotent.
ii.) f is nilpotent ⇐⇒ a0 , a1 , . . . , an are nilpotent.
iii.) f is a zero-divisor ⇐⇒ there exists a 6= 0 in A such that af = 0.
iv.) f is said to be primitive if (a0 , a1 , . . . , an ) = (1). Prove that if f, g ∈ A[x],
then f g is primitive ⇐⇒ f and g are primitive.
Pm
i.) ⇒: Suppose that f is a unit with inverse g = i=0 bi xi . Then since the
constant term in f g must be 1, we see that b0 a0 = 1 so a0 is a unit.
Further, by examining the highest order terms we get the relations:
an bm = 0
an−1 bm + an bm−1 = 0
an−2 bm + an−1 bm−1 + an bm−2 = 0
···
13
Multiplying the second equation by an we obtain
an−1 bm an + a2n bm−1 = 0
The first term vanishes from the first equation and we have a2n bm−1 = 0.
r+m
Inductively we can prove that ar+1
b0 = 0 so an
n bm−r = 0. Thus an
n
is nilpotent. Now note that −an x is nilpotent as well. It follows then
that a0 − an xn is a unit and we can argue as before to show that an−1 is
nilpotent and we are done.
Pn
⇐: Since a1 , . . . , an are nilpotent so is i=1 ai xi . (Just take a huge power
so that each turn vanishes). Thus f = a0 +nilpotent which is a unit by
problem 1.
ii.) ⇐: Clear
n
⇒: Suppose that f m = 0. Then am
n = 0 by expansion. Then f − an x
as a difference of two nilpotents is also nilpotent and we can repeat the
argument.
iii.) ⇐: extremely clear
Pm
⇒: Suppose that f g = 0 with g = i=0 bi xi of minimal degree. Then
an bm = 0. Also by commutativity and associativity, f (an g) = 0 and an g
has degree less than g. Thus we must have that an g = 0. But then this
means that an bk = 0 for every k. Thus f g = 0 really means:
0 = (a0 + · · · + an xn )(b0 + · · · + bm xm )
= (a0 + · · · + an−1 xn−1 )(b0 + · · · + bm xm )
Thus we argue as before to see that an−1 g = 0 and in P
general can do this
for an−k . So then just pick any nonzero bk and bk f = i bk ai = 0.
iv.) ⇒: Suppose that f g is primitive. Now suppose that f were not primitive.
Then (a0 , a1 , . . . , an ) 6= (1) so there is a common factor to all of the terms.
But then f g would have a common factor as well. Thus f is primitive, and
so must g be.
⇐: The classical Gauss Lemma, which is kind of fancy. INCOMPLETE
4. Show that in the ring A[x] the Jacobson Radical = Nilradical.
It is clear that since all maximal ideals are prime, that the Jacobson Radical
contains the Nilradical. Now suppose that f is in the Jacobson Radical. Then
1 − f g is a unit for every g ∈ A[x]. Let g = x. Then 1 − f x is a unit, so every
non constant term’s coefficient is nilpotent by exercise 2. But these are exactly
the original coefficients of f so f is nilpotent and thus in the nilradical.
6. A ring A is such that every ideal not contained in the nilradical contains
a non-zero idempotent (an element e with e2 = e 6= 0). Prove that the nilradical
and Jacobson radical of A are equal.
As in exercise 4, we only need to show that the Jacobson radical is contained
in the nilradical. Suppose that m is a maximal ideal that is not contained in
14
the nilradical. Then m has an idempotent, e 6= 0. Then since e is in a maximal
ideal, 1 − e is a unit. Finally,
1 − e = 1 − e2 = (1 − e)(1 + e).
Cancelling, we see that 1 + e = 1 and e = 0 which is a contradiction.
7. Let A be a ring in which every element x satisfies xn = x for some n > 1
(depending on x). Show that every prime ideal in A is maximal.
Let p be a prime ideal in A. Now let y be an element not in p. Then
y n − y = 0 ∈ p for some n. Thus y(y n−1 − 1) ∈ p and since p is prime, we
know that y n−1 − 1 ∈ p. Thus it follows that y n−1 − 1 = x for some x ∈ p and
p + (y) = (1) so that p is maximal.
8.Let A be a ring 6= 0. Show that the set of prime ideals of A has minimal
elements with respect to inclusion.
We first recall what a minimal element is. A minimal element is an element
m such that m ⊇ n implies m = n. We will fancifully use Zorn’s Lemma.
Suppose we have a chain of prime ideals p1 ⊇ p2 ⊇ · · · . Then consider their
intersection P = ∩∞
i=1 pi . This is clearly an ideal, and we now show that it is
prime. Suppose that xy ∈ P but that x, y ∈
/ P . Then there exists some i, j such
that x ∈
/ pi and y ∈
/ pj . Since xy ∈ P then it must happen that x ∈ pj and
y ∈ pi . WLOG, assume i > j. Then x ∈ pj ⊆ pi which is a contradiction. Thus
this chain has a lower bound and by Zorn’s lemma, has a minimal element.
9. Let a be an ideal 6= (1) in a ring A. Show that a = r(a) ⇐⇒ a is an
intersection of prime ideals.
⇒: Recall that r(a) is the preimage of the nilradical of A/a, which is an intersection of the prime ideals of A/a. This preimage is therefore the intersection
of all prime ideals that contain a
⇐: a is always contained in r(a). Suppose a isT an intersection of prime
ideals. Let x ∈ r(a). Then for some n, xn ∈ a = pi . The primeness of pi
guarantees that x ∈ a.
10. Let A be a ring, R its nilradical. Show that the following are equivalent:
i) A has exactly one prime ideal;
ii) every element of A is either a unit or nilpotent;
iii) A/R is a field.
i) ⇒ ii): Let x be a nonunit. Then it is contained in some maximal ideal
which must necessarily be R so x is nilpotent.
ii) ⇒ iii): Let x + R be an element of A/R. Then if x is nilpotent, it is the
0 element, and if not, then it is a unit so A/R is a field.
iii) ⇒ i): If A/R is a field, then R is maximal. Let p1 , p2 be two prime
ideals of A. Then p1 ! R which contradicts that R is maximal.
11. A ring A is Boolean if x2 = x for all x ∈ A. In a Boolean ring A, show
that
i) 2x = 0 for all x ∈ A.
15
ii) every prime ideal p is maximal, and A/p is a field with two elements.
iii) every finitely generated ideal in A is principal.
i): (x + 1)2 = x + 1 =⇒ x2 + 2x + 1 = x + 1 =⇒ x + 2x + 1 = x + 1 =⇒
2x = 0.
ii): The first part was shown in exercise 7. Let a + p be an element of A/p
and a ∈
/ p. This element is invertible since A/p is a field. Then since a2 = a we
have (a + p)2 = a + p. This implies (a + p) = 1 + p and thus the field has only
two elements.
iii): We claim that (x, y) = (xy + x + y). This is clear since x(xy + x + y) =
xy +x+xy = 2xy +x = x and similarly for y. The result follows from induction.
12. A local ring contains no idempotent 6= 0, 1.
If x ∈ A is not contained in the maximal ideal, then it is a unit and therefore
if it is idempotent, it is 1. Suppose that x2 = x for some x in the maximal ideal
m of A. Then 1 − x is not in m so it is a unit. But then
(1 − x)2 = 1 − 2x + x2 = 1 − x
which implies that 1 − x = 1 or that x = 0.
13. Let K be a field and let Σ be the set of all irreducible monic polynomials
f in one indeterminate with coefficients in K. Let A be the polynomial ring
over K generated by indeterminates xf , one for each f ∈ Σ. Let a be the ideal
of A generated by the polynomials f (xf ) for all f ∈ Σ. Show that a 6= (1).
Suppose that some combination of f (xf ) is equal to 1. Then for some polynomials gf in A we have
X
gf f (xf ) = 1
f ∈Σ
It is easy to convince yourself that this cannot be the case by examining what
would have to cancel out on the left side. (Of course this isn’t very satisfactory.
More to come later).
Let m be a maximal ideal of A containing a, and let K1 = A/m. Then K1
is an extension field of K in which each f ∈ Σ has a root. Repeat
S∞ this process
with K1 instead of K to get a new field K2 , and so on. Let L = n=1 Kn . Then
L is a field in which each f ∈ Σ splits completely into linear factors. Let K be
the set of all elements in L which are algebraic over K. Then K is the algebraic
closure of K.
14. In a ring A, let Σ be the set of all ideals in which every element is a
zero-divisor. Show that the set Σ has maximal elements and that every maximal
element of Σ is a prime ideal. Hence the set of zero-divisors in A is a union of
prime ideals
Order Σ by inclusion and let T be a chain. Then it is easy to see that the
union of all elements in the chain is in Σ and that it is an upper bound. Now
just apply Zorn’s Lemma.
Let m be a maximal element. Then we show it is prime. Suppose that
xy ∈ m, but x, y ∈
/ m. Since m is maximal, we need to have m + (x) and m + (x)
16
to have nonzero divisors. Suppose that m1 + ax ∈ m + (x) and m2 + by ∈ m + (y)
are not zero divisors. Then their product is not a zero divisor, and cannot be
in m. But look:
(m1 + ax)(m2 + by) = m1 (m2 + by) + m2 (ax) + abxy
which is an element of m. This is a contradiction.
The prime spectrum of a ring
15. Let A be a ring and let X be the set of all prime ideals of A. For each
subset E of A, let V (E) denote the set of all prime ideals of A which contain
E. Prove that
i) if a is the ideal generated by E, then V (E) = V (a) = V (r(a)).
ii) V (0) = X, V (1) = ∅.
iii) if (Ei )i∈I is any family of subsets of A, then
!
[
\
V
Ei =
V (Ei ).
i∈I
iv) V (a
T
b) = V (ab) = V (a)
S
i∈I
V (b) for any ideals a, b of A.
These results show that the sets V (E) satisfy the axioms for closed sets in a
topological space. The resulting topology is called the Zariski Topology. The
topological space X is called the prime spectrum of A, and is written Spec(A).
i): V (E) = V (a) is obvious. Since r(a) ⊇ a we have V (r(a)) ⊆ V (a). Finally,
suppose p is a prime ideal containing a. Then let x ∈ r(a). We will show x ∈ p
so that p contains r(a). xn ∈ a ⊆ p for some n, and since p is prime, x ∈ p.
ii): Clear
iii): Let
T p contain the union. Then in particular, p contains Ei for each i.
Thus p ∈ V (Ei ). Arguing in reverse easily proves the other direction.
iv): The first T
equality:
T
Since
ab
⊆
a
b, V (ab) ⊇ V (a b). Conversely, if p contains ab let x ∈
T
a b. Then x2 ∈ ab ⊆ p and since p is prime, x ∈ p.
The second equality:
⊇: Let p contain both a and b. Then it must contain the smaller ideal ab.
⊆: Let p contain ab. If p contains a we are done. So suppose not. Then let
y ∈ b, x ∈ a (x ∈
/ p). Now we have xy ∈ ab ⊆ p and since x ∈
/ p and p is prime
we must have y ∈ p so that p contains b.
16. Draw pictures of Spec(Z), Spec(R), Spec(C[x]), Spec(R[x]), Spec(Z[x]).
5
5.1
Modules
Modules and Module Homomorphisms
Let A be a commutative ring. Then an A-module is an abelian group M written
additively on which A acts linearly. What this means is that it is a pair (M, µ)
17
where M is an abelian group and µ : A × M → M such that if we write ax for
µ(a, x) (a ∈ A, x ∈ M ), the following axioms are satisfied for all a, b ∈ A and
all x, y ∈ M .
a(x + y) = ax + ay
(a + b)x = ax + bx
(ab)x = a(bx)
1x = x
All we are really concerned with the module structure however, is how multiplying by an element of the ring changes moves elements of M . In effect
for each a ∈ A there is a corresponding endomorphism of M , µa given by
µa (x) = ax. Thus to completely understand the module structure, we need
to know the function A → End(M ) sending a to µa . It is easy to see that
the above properties make this map into a ring homomorphism and even more
succinctly: An A-module is an abelian group M together with a ring homomorphism A → End(M ).
Example 5.1. 1) Any ideal of A is an A-module. In particular, A itself is an
A-module.
2) If A is a field k then an A-module is precisely a k-vector space.
3) If A = Z then a Z-module is the same as an abelian group. (Just define
the action nx = x + · · · + x). Question for prof. Polini. Every Z-module is an
abelian group, but must it have the trivial ring action?
4) If A = k[x] where k is a field, then an A-module is a k-vector space v with a
linear transformation T . Define µf : V → V by µf (v) = f (T )(v).
Let M and N be A-modules. We say that a function f : M → N is an
A-module homomorphism if f (x + y) = f (x) + f (y) and f (ax) = af (x). If A
is a field, then these are just the properties of a linear transformation. Note
that the composition of A-module homomorphisms is again an A-module homomorphism. Thus by defining addition and multiplication in an obvious way,
we can turn the set of all A-module homomorphisms from M to N into an
A-module. This A-module is denoted HomA (M, N ). Sometimes we might just
write Hom(M, N ) and you’ll have to guess what the ring is.
Now is where the fun starts. Suppose all the capital letters are A-modules,
and we have homomorphisms u : M 0 → M and v : N → N 00 . Then these induce
maps:
u : Hom(M, N ) → Hom(M 0 , N )
and
v : Hom(M, N ) → Hom(M, N 00 )
defined in exactly how you’d think they were:
u(f ) = f ◦ u
v(f ) = v ◦ f.
To illustrate this, consider the following commutative diagram: (I admit it
commutes quite trivially, but it still commutes)
Notice that there is a natural isomorphism Hom(A, M ) ∼
= M . Indeed, any
module homomorphism f : A → M is determined exactly by f (1) which can be
any element of M .
18
5.2
Submodules and Quotient Modules
A submodule M 0 of M is a subgroup of M which is closed under multiplication by
elements of A. The abelian group M/M 0 then inherits an A-module structure
by a(x + M 0 ) = ax + M 0 . This is well defined, and as for ideals, there is a
one-to-one order preserving correspondence between submodules of M/M 0 and
submodules of M which contain M 0 .
The kernel of a module homomorphism is the set of all x ∈ M such that
f (x) = 0. It is a submodule of M . The image of f is the set of all f (x) ∈ N
with x ∈ M . The cokernel of f is
Coker(f ) = N/Im(f ).
If M 0 is a submodule of M and M 0 ⊆ ker(f ) then we have an induced map.
f : M/M 0 → N
f (x + M 0 ) = f (x).
This is well defined (you should check this) and in particular if M 0 = kerf
then by the 1st isomorphism theorem, (or explicit calculation, we have that
M/kerf ∼
= Im(f )
We define the sum and intersection of modules in the same way we did for
rings.
Proposition 5.1. If L ⊇ M ⊇ N then
(L/N )/(M/N ) ∼
= L/M.
Proof. Consider the map f : L/N → L/M , f (x + N ) = x + M . This is well
defined (check if you don’t believe this!) and has kernel M/N so by the 1st
isomorphism theorem, we have the result.
Proposition 5.2. If M1 , M2 are submodules of M then
(M1 + M2 )/M1 ∼
= M2 /(M1 ∩ M2 ).
Proof. Consider the composite map
M2 → M1 + M2 → (M1 + M2 )/M1 .
This map is surjective (think about it). The kernel of the composite is M1 ∩ M2
and the result follows.
If a ⊂ A is an ideal then aM is exactly what you think it is, and is a
submodule of M .
We define (N : P ) = {a ∈ A : aP ⊂ N }. This is an ideal of A. In particular,
(0 : M ) = {a : aM = 0} is called the annihilator of M and we write Ann(M ).
This gives rise to a curious result: If a ⊆ Ann(M ) then M is an A/a module.
Indeed, if b + a ∈ A/a then define (b + a)x = bx. This is well-defined: Suppose
that b + a = c + a then b − c ∈ a implies (b − c)x = 0 for every x ∈ M and that
bx = cx.
We say that an A-module is faithful if Ann(M ) = 0.
19